Consider the nuclear reaction: (2/1)H + (9/4)Be --> (7/3)Li + (4/2) He. (A) How

Question

Consider the nuclear reaction:

(2/1)H + (9/4)Be --> (7/3)Li + (4/2) He.

(A) How much energy is liberated? (answer in units MeV)

(B) What is the threshold energy for this reaction? (answer in units MeV)

For part A, I tried to calculate the answer using the formula (Mproducts - Mreactants) * c^2 but I keep getting the wrong answer. I'm not sure where I am going wrong! Your help would be greatly appreciated!

I will rate lifesaver for the first person to help me answer the question correctly.

Thank you in advance!

Explanation / Answer

mass of deutrium = 2.014 u

mass of Be=9.0121822 u

mass of Li= 6.015 u

mass of He=4.0026 u

mass on left side = 2.014 + 9.0121822 = 11.0261822 u

mass on right side = 6.015 + 4.0026 = 10.0176 u

missing mass = 11.0261822-10.0176 = 1.0085822 u

energy liberated = 1.0085822*931 MeV = 938.9900282 MeV

threshold energy = (10.0176 + 11.0261822)*(1.0085822)*931/(2*9.0121822) = 1096.288 MeV

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